[Haskell-cafe] Puzzled by error with forall quantifier.

[Theodore S. Norvell]

Iavor S. Diatchki wrote:

> just skip the forall.   "free" type variables are implicitly 
> forall-quantified.

Of course!  Thanks.  The problem is that I actually wanted "exists".
However as the type that exists is functionally dependent on other the
parameters to the class I think I can deal with it using functional
dependencies, as you suggest.

Sorry for the elementary question.

Cheers,
Theo Norvell